Two corners of a triangle have angles of $\frac{7 π}{12}$ $(7 pi ) / 12$ and $\frac{π}{12}$ $pi / 12$ . If one side of the triangle has a length of $7$ $7$ , what is the longest possible perimeter of the triangle?

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Answer 1 · 2017-12-08T14:58:22

Largest possible area of the triangle is 79.1852

Given are the two angles $\frac{7 π}{12}$ $(7pi)/12$ and $\frac{π}{12}$ $pi/12$ and the length 7

The remaining angle:

$= π - (\frac{7 π}{12}) + \frac{π}{12} = \frac{π}{3}$ $= pi - ((7pi)/12) + pi/12 = (pi)/3$

I am assuming that length AB (7) is opposite the smallest angle.

Using the ASA

Area $= \frac{c^{2} \cdot sin (A) \cdot sin (B)}{2 \cdot sin (C)}$ $=(c^2*sin(A)*sin(B))/(2*sin(C)$

Area $= \frac{7^{2} \cdot sin (\frac{π}{3}) \cdot sin (\frac{7 π}{12})}{2 \cdot sin (\frac{π}{12})}$ $=( 7^2*sin(pi/3)*sin((7pi)/12))/(2*sin(pi/12))$

Area $= 79.1852$ $=79.1852$

Two corners of a triangle have angles of (7 pi ) / 12 7π12(7 pi ) / 12 and pi / 12 π12 pi / 12 . If one side of the triangle has a length of 7 77 , what is the longest possible perimeter of the triangle?